2. EENG223 Mesh Analysıs
2
Mesh Analysis
z Nodal analysis was developed by applying
KCL at each non-reference node.
z Mesh analysis is developed by applying KVL
around meshes/loops in the circuit.
z Mesh analysis results in a system of linear
equations which must be solved for unknown
currents.
3. EENG223 Mesh Analysıs
3
Mesh Analysis
z quantity of interest is current
z a mesh is a loop that does not contain
another loop within it
z work for planar circuit only
planar circuit -> no branch passes over or
under other branch
z M-meshes -> assign clockwise current for
each mesh
z apply KVL around each mesh
5. EENG223 Mesh Analysıs
5
Steps of Mesh Analysis
1. Identify meshes.
2. Assign a current to each mesh.
3. Apply KVL around each loop to get an
equation in terms of the loop currents.
4. Solve the resulting system of linear
equations.
7. EENG223 Mesh Analysıs
7
Steps of Mesh Analysis
1. Identify mesh (loops).
2. Assign a current to each mesh.
3. Apply KVL around each loop to get an
equation in terms of the loop currents.
4. Solve the resulting system of linear
equations.
9. EENG223 Mesh Analysıs
9
Steps of Mesh Analysis
1. Identify mesh (loops).
2. Assign a current to each mesh.
3. Apply KVL around each loop to get an
equation in terms of the loop currents.
4. Solve the resulting system of linear
equations.
13. EENG223 Mesh Analysıs
13
Steps of Mesh Analysis
1. Identify mesh (loops).
2. Assign a current to each mesh.
3. Apply KVL around each loop to get an
equation in terms of the loop currents.
4. Solve the resulting system of linear
equations.
14. EENG223 Mesh Analysıs
14
Matrix Notation
z The two equations can be combined into a
single matrix/vector equation.
−
=
Ω
+
Ω
Ω
−
Ω
−
Ω
+
Ω
2
1
2
1
k
1
k
1
k
1
k
1
k
1
k
1
V
V
I
I
15. EENG223 Mesh Analysıs
15
Solving the Equations
Let: V1 = 7V and V2 = 4V
Results:
I1 = 3.33 mA
I2 = -0.33 mA
Finally
Vout = (I1 - I2) 1kΩ = 3.66V
19. EENG223 Mesh Analysıs
19
Current Sources
z The current sources in this circuit will have
whatever voltage is necessary to make the
current correct.
z We can’t use KVL around the loop because
we don’t know the voltage.
z What to do?
20. EEE 223 Mesh Analysıs
20
Current Sources
z The 4mA current source sets I2:
I2 = -4 mA
z The 2mA current source sets a constraint on
I1 and I3:
I1 - I3 = 2 mA
z We have two equations and three unknowns.
Where is the third equation?